The equation of a circle is given below. $(x-5)^{2}+(y+2)^{2} = \dfrac{81}{4}$ What is its center? $($
Solution: Standard equation of the circle A circle is the collection of all points at a distance ${r}$ from a center $({h},{k})$. We can use the Pythagorean theorem to write an equation to relate the center and radius. $({h},{k})$ ${r}$ $x-{h}$ $y-{k}$ $(x, y)$ $\begin{aligned} a^2+b^2&=c^2\\\\ (x - {h})^2 + (y - {k})^2 &= {r}^2 \end{aligned}$ Rewriting the given equation We can rewrite the given equation as: $\begin{aligned}(x-5)^{2}+(y+2)^{2} &= \dfrac{81}{4}\\\\ (x - {5})^2 + (y - {(-2)})^2 &= { \dfrac{81}{4}}\end{aligned}$ Finding the center According to the rewritten equation, we can see that the center of the circle is $({5}, {-2})$. Finding the radius According to the standard equation of the circle, we get that ${r^2}={ \dfrac{81}{4}}$. Solving for the radius, we get that $r={\sqrt{ \dfrac{81}{4}}}= {\dfrac92}$. Summary The circle is centered at $(5, -2)$. The circle has a radius of $\dfrac{9}{2}$ units.